Mbpp/296

def get_Inv_Count(arr):
    """
    Count inversions in the given array.
    An inversion is a pair (i, j) such that i < j and arr[i] > arr[j].
    This implementation uses a merge-sort based approach with O(n log n) time complexity.
    """
    def merge_and_count(left, right):
        merged = []
        count = 0
        i = j = 0
        while i < len(left) and j < len(right):
            if left[i] <= right[j]:
                merged.append(left[i])
                i += 1
            else:
                # All remaining elements in left are greater than right[j]
                merged.append(right[j])
                count += len(left) - i
                j += 1
        # Append remaining elements
        merged.extend(left[i:])
        merged.extend(right[j:])
        return merged, count

    def sort_and_count(arr):
        if len(arr) <= 1:
            return arr, 0
        mid = len(arr) // 2
        left, left_count = sort_and_count(arr[:mid])
        right, right_count = sort_and_count(arr[mid:])
        merged, merge_count

• [[1, 20, 6, 4, 5]]

• [[10, 9, 8, 7, 6, 5, 4, 3, 2, 1]]